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A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

My work:

The restoring force is $-kx$. Then

$$7.5 = -k(2.5) \\ -\frac{7.5}{2.5} = k \\ ma = -\frac{7.5x}{2.5} \\ my’’ + 9y’ + -3y = 0,\quad y(0) = 2.5, y(5) = 0 \\ \frac{-9 \pm \sqrt{81 + 4(m)(3)}}{2m} \\ -\frac{9}{2m} \pm \frac{\sqrt{81+12m}}{2m} \\ y = Ae^{-(9/2)x}\cos\left(\frac{\sqrt{81+12m}}{2m}x\right) + Be^{-(9/2)x}\sin\left(\frac{\sqrt{81+12m}}{2m}x\right) \\ 2.5 = A + B\cdot 0 \\ 0 = (2.5)e^{-45/2}\cos\left(\sqrt{81+12m}\frac{5}{2m}\right) + Be^{-45/2}\sin\left(\sqrt{81+12m}\frac{5}{2m}\right)$$

Any help would be appreciated

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  • $\begingroup$ Not sure of my edit. Since user115277 isn't still around, if someone else wants to go through this and make sure I edited correctly, it'd be helpful. $\endgroup$ – user137731 Nov 26 '16 at 22:46
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1) find the $k$

2)$\zeta=c/(2\sqrt{k/m})=1 $ for critical damping. So solve for $m$ as you alreaday know $c$ and $k$.

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A short, non insightful answer:

$$ \omega_0^2 = \frac{k}{m} $$ $$\alpha=\frac{c}{2m}$$

If the system is critically damped, $\omega_0^2 = \alpha^2$, so: $$\frac{3}{m}=\frac{81}{4m^2} $$ $$m=6.75kg$$

If the system is overdamped damped, $\omega_0^2 \lt \alpha^2$

If the system is underdamped damped, $\omega_0^2 \gt \alpha^2$

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